3. 79.0 grams of water was produced with a percent yield of 75.0%. How many grams of oxygen did he start with?

4. 2.00 moles of oxygen gas at STP will produce how many grams of water, if the percent
yield is 82.0%?

Explanation:

3 )

18 grams of water contains 16 grams of oxygen .

Let the grams of oxygen needed be m .

yield is 75%

Only 75 % of m will make water .

.75 m will form water .

water formed by .75 m of oxygen = 18 / 16 x .75 m

Given that

18 / 16 x .75 m = 79

m = (16 x 79) / ( 18 x .75 )

= 93.63 grams .

So oxygen required is 93.63 grams .

4 )

One mole of water contains 0.5 moles of oxygen.

2 moles of oxygen will make 2 /,5 = 4 moles of water .

percent yield is 82 % ,

so water produced = .82 x 4 = 3.28 moles of water.

the answer is c i believe